# Front-door adjustment formula: difficulty in reconcile the two formula

In The Book Of Why, the author gives two formulas related to the front-door adjustment formula.

On page 227 (formula 7.1), the front-door adjustment is given by $$P(Y|do(X)) = \sum_zP(Z=z|X)\sum_xP(Y|X=x,Z=z)P(X=x)$$

On page 236 (figure 7.4), the front-door adjustment is given by $$P(c|do(s)) = \sum_{s'}\sum_tP(c|t,s')P(s')P(t|s)$$ Or, after aligning the notation $$P(Y|do(X)) = \sum_{X'}\sum_ZP(Y|z,x')P(x')P(z|x)$$

I have read this question, and understand that the final formula can be expressed as: $$P(Y|do(X)) = \sum_Z\sum_{X'}P(z|x)P(Y|z,x')P(x')$$ which is the same as formula 7.1.

But I am not sure why $$x$$ becomes $$x'$$ in the last formula?

and secondly since $$\sum_{X'}\sum_ZP(Y|z,x')P(x')P(z|x)$$ is actually a multiplication inside, can I write it as $$\sum_Z\sum_{X'}P(Y|z,x')P(x')P(z|x)$$?

• x' is the intrevention observed in the data, x is the intervention of interest, the do(x). See also Derivation of front door adjustment without intervention on the mediator doi.org/10.6084/m9.figshare.20278347.v1 Commented Jul 9, 2022 at 9:53

Question 1: Why does $$x$$ becomes $$x'$$ in the last formula?
Answer: $$x$$ does not "become" $$x'$$ in the last formula. $$x'$$ is a variable of summation, and hence a "dummy" variable. It has no scope outside the summation. You could just as easily write $$P(Y|\operatorname{do}(X)) = \sum_Z\sum_{X}P(z|x)P(Y|z,\xi)P(\xi).$$ On the other hand, the $$x$$ shows up on both sides of the equation, and definitely has scope inside and outside the summation. So the $$x$$ is a particular value of the random variable $$X,$$ whereas the $$x'$$ is a dummy variable that ranges over all the possible values of $$X.$$ Here's what I would regard as the clearest way to write the equation: $$P(Y=y|\operatorname{do}(X=x)) = \sum_{Z=z}\sum_{X=x'}P(Z=z|X=x)P(Y=y|Z=z,X=x')P(X=x').$$ This is clunkier, I admit.
Question 2: ... since $$\sum_{X'}\sum_ZP(Y|z,x')P(x')P(z|x)$$ is actually a multiplication inside, can I write it as $$\sum_Z\sum_{X'}P(Y|z,x')P(x')P(z|x)$$?